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    Rolling a die 600 times




    So I was given a question that states that "A gambler buys a new die and throws it 600 times. Assuming the die is unbiased, estimate the probability that he obtains between 90 and 100 "sixes". What might he conclude if he were to have obtained 120 "sixes"?"

    A few people in my class suggested calculating the z-scores and finding the probability using the normal distribution. However I didn't think this was possible given that we weren't given a standard distribution or anyway that I could perceive for calculating the mean.

    Instead I had said that rolling a 6 was a 'success' with p=1/6 and rolling any other number was a 'failure' and that this was a binomial distribution, specifically the Poisson distribution since n is large and p is small. For lambda I had lambda = np =600(1/6) =100. Then i calculated Pr(90<X<100) = Pr(X=100) - Pr(X=90)
    (using Pr(X=r) = lambda^r/r! (e^-lambda) for poisson I had) = 100^100/100! (e^-100) - 100^90/90! (e^-9). I calculated Pr(X=120) similiarly and said it was a very unlikely occurrence.

    I have this question due soon and I'm feeling a bit lost. Any help at all would be greatly appreciated

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    Re: Rolling a die 600 times

    For 600 throws, I would use a normal approximation with μ = (1/6)600 = 100 and σ = (1/6)(1–1/6)600 = 250/3 to calculate P(89.5 < X < 100.5) and P(119.5 < X < 120.5), especially since these values are quite close to the expected mean μ = 100.

    Of course, you could use an infinite-precision package like Matlab or Mathematica to calculate the exact answer.
    Last edited by Con-Tester; 04-01-2015 at 12:17 PM. Reason: Typing failure...

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    Re: Rolling a die 600 times


    Quote Originally Posted by kittykat123 View Post
    . For lambda I had lambda = np =600(1/6) =100. Then i calculated Pr(90<X<100) = Pr(X=100) - Pr(X=90)
    (using Pr(X=r) = lambda^r/r! (e^-lambda) for poisson I had) = 100^100/100! (e^-100) - 100^90/90! (e^-9). I calculated Pr(X=120) similiarly and said it was a very unlikely occurrence.
    Apparently you didn't use cumulative sums. Concerning the 120, the question is what interval to sum over, for one thing. If you sum from 120 upward ("to infinity") what exactly have you found? Simpler to sum from 80 to 120 and subtract that result from
    1 to have the probability of rolling 120 (or 80) sixes in 600 rolls.

    Art

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